Let f_kx=1/ksin^(k )x+cos^(k )x where x∈R and k≥1. Then f_4x-f_6x equals

Correct answer: (B) 1/(1 2). f_4x=1/4sin^4x+cos^4x =1/4sin2x+cos2x^2-2sin^2x cos^2x =1/41-2sin^2x cos^2x f_6x=1/6sin^6x+cos^6x =1/6sin2x+cos2x^3-3sin^2x cos^2xsin^2x+cos^2x =1/61-3sin^2x cos^2x1 f_4x-f_6x=1/4-1/6 =1/12

Q58 / 90 · JEE Main 2014 (06 Apr) · Mathematics

Let $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x)$ where $x \in R$ and $k \geq1$ . Then $f_{4} (x) - f_{6} (x)$ equals

A
$\frac{1}{4}$
Correct answer
B
$\frac{1}{12}$
C
$\frac{1}{6}$
D
$\frac{1}{3}$

Solution

f_{4} (x) = \frac{1}{4} (\sin^{4} x + \cos^{4} x)

= \frac{1}{4} [{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x]

= \frac{1}{4} [1 - 2 \sin^{2} x \cos^{2} x]

f_{6} (x) = \frac{1}{6} [\sin^{6} x + \cos^{6} x]

= \frac{1}{6} [{(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)]

= \frac{1}{6} [1 - 3 \sin^{2} x \cos^{2} x (1)]

f_{4} (x) - f_{6} (x) = \frac{1}{4} - \frac{1}{6}

= \frac{1}{12}

Hence, the correct answer is option: 'B'.

Let fkx=1ksink x+cosk x where x∈R and k≥1. Then f4x-f6x equals

Solution

Relevant Questions

Let $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x)$ where $x \in R$ and $k \geq1$ . Then $f_{4} (x) - f_{6} (x)$ equals